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hw11_sol

# hw11_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-9* Use the method of joints to determine the force in each member of the truss shown in Fig. P7—9. State whether each member is in tension or compression. SOLUTION For this simple truss, the _member forces can be determined without solving for the support reactions. From a free-body diagram for joint D: + ‘\ ZFy - TED sin 30° — 1000 sin 60° = 0 Tan -l732.05 lb 8 1732 lb (C) Ans. 0 0 + x” EFX -TCD - TED cos 30 - 1000 cos 60 -TCD — (-1732.05) cos 30° - 1000 cos 60° = 0 TCD 1000 1b = 1000 lb (T) Ans. From a free-body diagram for joint C: + /” EFx = -TAC + TCD = -TAC + 1000 s 0 TAC = 1000 1b = 1000 lb (T) Ans. + '\ EFY 'Tac = o AnS- From a free-body diagram for joint B: 0 0 + -4 2Fx - —TAB - TBC cos 60 + TBD cos 60 -TAB - (0) + (-1732.05) cos 60° -866.03 lb 3 866 1b (C) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Use the method of joints to determine the force in each member of the truss shown in Fig. P7-11. State whether each member is in tension or compression. Fig. P7-11 SOLUTION For this simple truss, the member forces can be p determined without solving for the support reactions. From a free-body diagram for joint B: + /” 2F = T cos 30° - 750 cos 30° x AB TAB 750 lb = 750 lb (T) Ans. . , O - - . . 0 + R\ EFY -TBC — TAB Sln 30 130 Sln JO -ch - (750) sin 30° - 750 sin 30° = 0 TBC = —750 lb = 750 lb (C) Ans. From a free-body diagram for Joint C: + /” 2F = r - 900 cos 30° = 0 x AC : 779.4 lb 2 779 lb (T) . _ ' 0 CD 900 Sln 30 __,_ . O 'IDU - TCD - 900 Sin 30 — O = -1200 lb = 1200 lb (C) Ans. From a free-body diagram for Joint D: ., _. . o + T ZFY TAD + TCD sin 30 - TAD + ( 1200) Sln 30 , TAD = 600 lb 3 600 lb (T) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-13* Determine the forces in members AB, AE, and EF of the truss shown in Fig. P7-13. State whether each member is in tension or compression. ' I fur-13%}: win-km} «may! ~.' ' SOLUTION For this simple truss, the required member forces can be determined without solving for the support reactions. From a free-body diagram for joint A: + T 2F = T sin 30° — 600 = o y AB TAB = 1200 lb = 1200 lb (T) O + TAB cos 30 0 TAE + 1200 cos 30 — 0 — -1039.2 lb 2 1039 lb (C) From a free-body diagram for Joint B: + ‘\ spy = -TBE - 750 = o TBE = -750 lb = 750 1b (C) From a free-body diagram for joint E: _ _ 0- .v0 . o_ _ o + K\ ZFy — TBE sin 60 TEF sin 60 + TA: Sln 60 800 Sin 30 (-750) sin 60° - TEF sin 60° (-1039 2) sin 60° - 800 sin 30° = 0 -2251.1 lb 8 2250 lb (C) Ans. 1:135? ENGINEERING MECHANICS - STATICS‘ 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the forces in members AD, CD. anﬂ DE of the truss shown in Fig. P7-15. State whether each member is in tension or compression. SOLUTION From a free—body diagram for the complete truss: + C EMA = By(6) — 3000(0) — 2000(12) = By = 7000 lb = 7000 lb T From a free-body diagram for Joint B: +T2Fy:TBD+7000=0 TBD = -7000 lb = 7000 lb (C) From a free-body diagram for joint F: + T :5 = —T sin 45° - 2000 = 0 y DF -2828 lb 3 2830 lb (C) -0 --I'EF — I'DF cos 40 l -0 - --I'EF - (-d828) cos 4o - 0 2000 lb = 2000 lb (T) ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-15 (Continued) From a free-body diagram for joint E: 0 + —+ 2Fx - TEF - TCE cos 45 , 2000 - TCE cos 45 = 0 2828 1b = 2828 lb (T) . o _ -TCE sin 45 - TDE - 3000 - 0 -(2828) sin 45° - TDE - 3000 = o -5000 1b = 5000 lb (C) Ans. From a free-body diagram for joint D: DE DF AD '80 + T 2Fy = T + T sin 45° - T sin 45° - T = (-5000) + (-2828) sin 45° — T sin 45° — (-7000) = AD 0 Ans. 0 0 — TDF cos 45 - TCD - TAD cos 45 - 0 o — (-2828) cos 45 - TCD — 0 — 0 -2000 1b 3 2000 lb (C) 526 ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the forces in members BC, CD, and DF of the truss shown in Fig. P7-16. State whether each member is in tension or compression. SOLUTION From a free-body diagram for the complete truss: + C EMA = Ey(8) - 10(4) - 8(6) = 0 EV = 11.00 kN = 11.00 kN T From a free-body diagram for joint E: + T EF T sin 30° + 11.00 = 0 y DE TDE = -22.00 kN = 22.0 kN (C) SKN ’Is From a free—body diagram “‘ “X for joint D: T 3 I c I = —T cos 30° - 8 cos 30° = 0 D £5 OF I ~8.00 kN = 8.00 kN (C) Ans. “TD; ,’D \ ,gO,’ Tne‘x TDE - TCD + 8 sin 30° — TDF sin 30° (-22.00) — TCD + 8 sin 30° — (—8.00) sin 30° = 0 -14.00 kN = 14.00 kN (C) Ans. From a free-body diagram for joint C: O 0 + -+ ZFX - -TBC cos 30 A+ TCD cos 30 = -TBC cos 300 + (-14.00) cos 300 = 0 —14.00 kN = 14.00 kN (C) Ans. ...
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hw11_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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